//将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
//
// 
//
// 示例： 
//
// 输入：1->2->4, 1->3->4
//输出：1->1->2->3->4->4
// 
// Related Topics 链表 
// 👍 1374 👎 0


package leetcode.editor.cn;

public class MergeTwoSortedLists {
    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(2);
        l1.next.next = new ListNode(4);
        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);
        Solution solution = new MergeTwoSortedLists().new Solution();
        ListNode listNode = solution.mergeTwoLists(l1, l2);
        System.out.println(listNode);
    }
    //leetcode submit region begin(Prohibit modification and deletion)

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     * int val;
     * ListNode next;
     * ListNode() {}
     * ListNode(int val) { this.val = val; }
     * ListNode(int val, ListNode next) { this.val = val; this.next = next; }
     * }
     */
    class Solution {
        public ListNode mergeTwoLists(ListNode l1, ListNode l2) {

            //头指针
            ListNode head = new ListNode();
            //尾指针
            ListNode tail = new ListNode();
            head = tail;
            while (l1 != null && l2 != null) {
                if(l1.val <= l2.val){
                    tail.next = l1;
                    l1 = l1.next;
                }else {
                    tail.next = l2;
                    l2 = l2.next;
                }
                tail = tail.next;
            }

            if(l1!=null){
                tail.next = l1;
            }

            if(l2!=null){
                tail.next = l2;
            }
            return head.next;
        }
    }


//leetcode submit region end(Prohibit modification and deletion)
public static class ListNode {
    int val;
    ListNode next;

    ListNode() {
    }

    ListNode(int val) {
        this.val = val;
    }

    ListNode(int val, ListNode next) {
        this.val = val;
        this.next = next;
    }
}
}